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nata0808 [166]
3 years ago
12

An amusement park charges $27 for adult tickets (a) and $24 for children's tickets (c). Last Friday the amusement park sold 272

tickets and collected $7,137. Which of the following systems of equations is a correct representation of this data?
Chemistry
2 answers:
Vaselesa [24]3 years ago
7 0
Try using a calculator
Lapatulllka [165]3 years ago
4 0

Answer:

a+c=272

27a+24c=7,137

Explanation: Hope this helps!!!

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Determining what it is you want to know:
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2 years ago
A 34.53 ml sample of a solution of sulfuric acid, h2s04, reacts with 27.86 ml of 0.08964 m naoh solution. calculate the molarity
Gnoma [55]
The balanced equation between NaOH and H₂SO₄ is as follows
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
number of moles of NaOH moles reacted = molarity of NaOH x volume
number of NaOH moles = 0.08964 mol/L x 27.86 x 10⁻³ L = 2.497 x 10⁻³ mol
according to molar ratio of 2:1
2 mol of NaOH reacts with 1 mol of H₂SO₄
therefore 2.497 x 10⁻³ mol of NaOH reacts with - 1/2 x 2.497 x 10⁻³ mol of H₂SO₄
number of moles of H₂SO₄ reacted - 1.249 x 10⁻³ mol 
Number of H₂SO₄ moles in 34.53 mL - 1.249 x 10⁻³ mol 
number of H₂SO₄ moles in 1000 mL - 1.249 x 10⁻³ mol / 34.53 x 10⁻³ L = 0.03617 mol 
molarity of H₂SO₄ is 0.03617 M
6 0
3 years ago
When a 1.00 L sample of water from the surface of the Dead Sea (which is more than 400 meters below sea level and much saltier t
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Answer: Molarity of MgCl_2 in the original sample was 1.96M

Explanation:

Molarity is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{\text{no of moles}}{\text{Volume in L}}

{\text {moles of solute}=\frac{\text {given mass}}{\text {molar mass}}=\frac{186g}{95g/mol}=1.96

Now put all the given values in the formula of molarity, we get

Molarity=\frac{1.96}{1.00L}

Molarity=1.96mol/L

Thus molarity of MgCl_2 in the original sample was 1.96M

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3 years ago
Arrange the following oxides in order of increasing acidity.
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Answer:

Based on the Modern Periodic table, there is an increase in the electropositivity of the atom down the group as well as increases across a period. On comparing the electropositivities of the mentioned oxides central atom, it is seen that Ca is most electropositive followed by Al, Si, C, P, and S is the least electropositive.  

With the decrease in the electropositivity, there is an increase in the acidity of the oxides. Thus, the increasing order of the oxides from the least acidic to the most acidic is:  

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